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Lecture Presentation Chapter 6 Electronic Structure of Atoms © 2012 Pearson Education, Inc. John D. Bookstaver St. Charles Community College Cottleville, MO Waves • To understand the electronic structure of atoms, one must understand the nature of electromagnetic radiation. • The distance between corresponding points on adjacent waves is the wavelength (). © 2012 Pearson Education, Inc. Electronic Structure of Atoms Waves • The number of waves passing a given point per unit of time is the frequency (). • For waves traveling at the same velocity, the longer the wavelength, the smaller the frequency. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. B. C. D. 1.0 m and 3.0 108 cycles/s 2.0 m and 6.0 108 cycles/s 0.5 m and 3.0 108 cycles/s 0.5 m and 6.0 108 cycles/s Electronic Structure of Atoms Electromagnetic Radiation • All electromagnetic radiation travels at the same velocity: the speed of light (c), 3.00 108 m/s. • Therefore, c = © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. B. C. D. An X-ray has a longer wavelength and thus a higher frequency. An X-ray has a shorter wavelength and thus a higher frequency. An X-ray has a shorter wavelength and thus a lower frequency. An X-ray has a longer wavelength and thus a lower frequency. Electronic Structure of Atoms Sample Exercise 6.1 Concepts of Wavelength and Frequency Two electromagnetic waves are represented in the margin. (a) Which wave has the higher frequency? (b) If one wave represents visible light and the other represents infrared radiation, which wave is which? Solution Practice Exercise If one of the waves in the margin represents blue light and the other red light, which is which? Electronic Structure of Atoms Sample Exercise 6.2 Calculating Frequency from Wavelength The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation? Solution Practice Exercise (a) A laser used in eye surgery to fuse detached retinas produces radiation with a wavelength of 640.0 nm. Calculate the frequency of this radiation. (b) (b) An FM radio station broadcasts electromagnetic radiation at a frequency of 103.4 MHz (megahertz; 1 MHz = 106s-1 ). Calculate the wavelength of this radiation. The speed of light is 2.998 × 108 m/s to four significant digits. Electronic Structure of Atoms A. Yes: X-rays travel at at faster speeds than visible light. B. No: both X-rays and visible light travel at the speed of light, c. Electronic Structure of Atoms The Nature of Energy The wave nature of light does not explain how an object can glow when its temperature increases. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. White-yellow or yellow area in center, bottom and middle B. Blue-green area at right bottom C. Green-yellow at middle right D. Black area at left Electronic Structure of Atoms The Nature of Energy Max Planck explained it by assuming that energy comes in packets called quanta. © 2012 Pearson Education, Inc. Electronic Structure of Atoms The Nature of Energy • Einstein used this assumption to explain the photoelectric effect. • He concluded that energy is proportional to frequency: E = h where h is Planck’s constant, 6.626 10−34 J-s. © 2012 Pearson Education, Inc. Electronic Structure of Atoms The Nature of Energy • Therefore, if one knows the wavelength of light, one can calculate the energy in one photon, or packet, of that light: c = E = h © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. Lack of gas molecules keeps inside of vacuum tube clean. B. Electrons cannot move through a gas. C. Electrons would strike gas molecules and many electrons would not readily arrive at the positive terminal. D. Gas molecules in tube would prevent observation of any emission of electrons. Electronic Structure of Atoms A. B. C. D. 5 x 10-3 J 3 x 10-36 J 6 x 10-30 J 6.63 x 10-34 J Electronic Structure of Atoms Sample Exercise 6.3 Energy of a Photon Calculate the energy of one photon of yellow light that has a wavelength of 589 nm. Solution Practice Exercise (a) A laser emits light that has a frequency of 4.69 × 1014s-1. What is the energy of one photon of this radiation? (b) If the laser emits a pulse containing 5.0 × 1017 photons of this radiation, what is the total energy of that pulse? (c) If the laser emits 1.3 × 10-2J of energy during a pulse, how many photons are emitted? Electronic Structure of Atoms A. Photon of infrared light B. Photon of ultraviolet light Electronic Structure of Atoms The Nature of Energy Another mystery in the early twentieth century involved the emission spectra observed from energy emitted by atoms and molecules. © 2012 Pearson Education, Inc. Electronic Structure of Atoms The Nature of Energy • For atoms and molecules, one does not observe a continuous spectrum, as one gets from a white light source. • Only a line spectrum of discrete wavelengths is observed. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. Only certain orbits are allowed. B. An electron can only exist in one “allowed” energy state. C. Electrons do not spiral into the nucleus. D. Electrons can only change from one “allowed” energy state to another. Electronic Structure of Atoms The Nature of Energy • Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: 1. Electrons in an atom can only occupy certain orbits (corresponding to certain energies). 2. Electrons in permitted orbits have specific, “allowed” energies; these energies will not be radiated from the atom. 3. Energy is only absorbed or emitted in such a way as to move an electron from one “allowed” energy state to another; the energy is defined by E = h © 2012 Pearson Education, Inc. Electronic Structure of Atoms The Nature of Energy The energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation: E = −hcRH ( 1 1 n f2 ni2 ) where RH is the Rydberg constant, 1.097 107 m−1, and ni and nf are the initial and final energy levels of the electron. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. Ultraviolet light, because a greater amount of energy is emitted during the second electron transition than for the one with emission of visible light B. Infrared light, because a greater amount of energy is emitted during the second electron transition than for the one with emission of visible light C. Ultraviolet light, because a smaller amount of energy is emitted during the second electron transition than for the one with emission of visible light D. Infrared light, because a smaller amount of energy is emitted during the second electron transition than for the one with emission of visible light Electronic Structure of Atoms A. It neither emits nor absorbs energy. B. It both emits and absorbs energy simultaneously. C. It emits energy. D. It absorbs energy. Electronic Structure of Atoms Sample Exercise 6.4 Electronic Transitions in the Hydrogen Atom Using Figure 6.12, predict which of these electronic transitions produces the spectral line having the longest wavelength: n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3. Solution Practice Exercise Indicate whether each of the following electronic transitions emits energy or requires the absorption of energy: (a) n = 3 to n = 1; (b) n = 2 to n = 4. FIGURE 6.12 Energy states in the hydrogen atom. Only states for n = 1 through n = 4 and n = ∞ are shown. Energy is released or absorbed when an electron moves from one energy state to another. Electronic Structure of Atoms The Wave Nature of Matter • Louis de Broglie posited that if light can have material properties, matter should exhibit wave properties. • He demonstrated that the relationship between mass and wavelength was h = mv © 2012 Pearson Education, Inc. Electronic Structure of Atoms Sample Exercise 6.5 Matter Waves What is the wavelength of an electron moving with a speed of 5.97 × 106 m/s? The mass of the electron is 9.11 × 10-31 kg. Solution Practice Exercise Calculate the velocity of a neutron whose de Broglie wavelength is 500 pm. The mass of a neutron is given in the table inside the back cover of the text. Electronic Structure of Atoms A. Yes, but not observable because the matter wavelength is too small for our eyes to detect. B. Yes and observable because matter waves of macroscopic objects are in the visible region of light. C. Matter waves are not produced for macroscopic objects and thus none are observable. Electronic Structure of Atoms The Uncertainty Principle Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely is its position is known: (x) (mv) h 4 In many cases, our uncertainty of the whereabouts of an electron is greater than the size of the atom itself! © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. B. C. D. The charge of subatomic particles compared to zero charge for macroscopic particles. The small mass and size of subatomic particles compared to the mass and size of macroscopic particles. The small volume occupied by subatomic particles compared to the volume occupied by macroscopic particles. The slower speeds of subatomic particles Electronic compared to the speeds of macroscopic particles. Structure of Atoms Quantum Mechanics • Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated. • This is known as quantum mechanics. • The wave equation is designated with a lowercase Greek psi (). • The square of the wave equation, 2, gives a probability density map of where an electron has a certain statistical likelihood of being at any given instant in time. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. B. C. D. Far from the nucleus Along the z axis Along the y axis Near the nucleus Electronic Structure of Atoms Quantum Numbers • Solving the wave equation gives a set of wave functions, or orbitals, and their corresponding energies. • Each orbital describes a spatial distribution of electron density. • An orbital is described by a set of three quantum numbers. © 2012 Pearson Education, Inc. Electronic Structure of Atoms Principal Quantum Number (n) • The principal quantum number, n, describes the energy level on which the orbital resides. • The values of n are integers ≥ 1. © 2012 Pearson Education, Inc. Electronic Structure of Atoms Angular Momentum Quantum Number (l) • This quantum number defines the shape of the orbital. • Allowed values of l are integers ranging from 0 to n − 1. • We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals. Electronic © 2012 Pearson Education, Inc. Structure of Atoms Angular Momentum Quantum Number (l) Value of l 0 1 2 3 Type of orbital s p d f © 2012 Pearson Education, Inc. Electronic Structure of Atoms Magnetic Quantum Number (ml) • The magnetic quantum number describes the three-dimensional orientation of the orbital. • Allowed values of ml are integers ranging from −l to l: −l ≤ ml ≤ l • Therefore, on any given energy level, there can be up to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, and so forth. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. B. C. D. An orbital is composed of some integral number of orbits. An orbit is a well-defined circular path around the nucleus while an orbital is a wave function that gives the probability of finding the electron at any point in space. An orbit is a well-defined circular path around the nucleus while an orbital is the object (electron) that is moving around the nucleus. There is no difference between the definitions of the terms “orbit” and “orbital.” They simply were proposed Electronic by different scientists. Structure of Atoms Magnetic Quantum Number (ml) • Orbitals with the same value of n form a shell. • Different orbital types within a shell are subshells. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. B. C. D. One subshell labeled 4s Two subshells labeled 4s and 4p Three subshells labeled 4s, 4p and 4d Four subshell labeled 4s, 4p, 4d and 4f Electronic Structure of Atoms A. B. C. D. Because –1/(2)2 and –1/(1)2 are of much greater difference than –1/(3)2 and –1/(2)2 in Bohr’s equation that describes the hydrogen atom. Because the Rydberg in Bohr’s equation constant is higher for n = 1 to n = 2 transitions. Because the energy difference between the 1st and 2nd energy level of hydrogen is smaller than the energy difference between the 2nd and 3rd energy level. Because the electron in the hydrogen atom never Electronic Structure occupies energy levels of higher than n = 1. of Atoms Sample Exercise 6.6 Subshells of the Hydrogen Atom (a) Without referring to Table 6.2, predict the number of subshells in the fourth shell, that is, for n = 4. (b) Give the label for each of these subshells. (c) How many orbitals are in each of these subshells? Solution Practice Exercise (a) What is the designation for the subshell with n = 5 and l = 1? (b) How many orbitals are in this subshell? (c) Indicate the values of ml for each of these orbitals. Electronic Structure of Atoms s Orbitals • The value of l for s orbitals is 0. • They are spherical in shape. • The radius of the sphere increases with the value of n. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. B. C. D. Maxima 4 4 3 3 Nodes 4 3 3 4 Electronic Structure of Atoms s Orbitals Observing a graph of probabilities of finding an electron versus distance from the nucleus, we see that s orbitals possess n − 1 nodes, or regions where there is 0 probability of finding an Electronic electron. © 2012 Pearson Education, Inc. Structure of Atoms p Orbitals • The value of l for p orbitals is 1. • They have two lobes with a node between them. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. Electrons are moving slower at edges than in interior of the lobes B. Greater number of electrons at edges than in interior of the lobes. C. Probability of finding an electron at edges is lower than in interior of the lobes D. Number of electrons is disappearing near the edges of the lobes Electronic Structure of Atoms A. B. C. D. 2px 2py 2pz 2p Electronic Structure of Atoms d Orbitals • The value of l for a d orbital is 2. • Four of the five d orbitals have 4 lobes; the other resembles a p orbital with a doughnut around the center. © 2012 Pearson Education, Inc. Electronic Structure of Atoms Energies of Orbitals • For a one-electron hydrogen atom, orbitals on the same energy level have the same energy. • That is, they are degenerate. • As the number of electrons increases, though, so does the repulsion between them. • Therefore, in many-electron atoms, orbitals on the same energy level are no longer degenerate. © 2012 Pearson Education, Inc. Electronic Structure of Atoms A. B. C. D. 4g 4d 4p and 4d 4d and 4f Electronic Structure of Atoms A. B. C. D. 3 6 9 18 Electronic Structure of Atoms A. B. C. D. 3s < 3p < 3d 3s < 3d < 3p 3d < 3p < 3s 3p < 3s < 3d Electronic Structure of Atoms Spin Quantum Number, ms • In the 1920s, it was discovered that two electrons in the same orbital do not have exactly the same energy. • The “spin” of an electron describes its magnetic field, which affects its energy. © 2012 Pearson Education, Inc. Electronic Structure of Atoms Spin Quantum Number, ms • This led to a fourth quantum number, the spin quantum number, ms. • The spin quantum number has only 2 allowed values: +1/2 and −1/2. © 2012 Pearson Education, Inc. Electronic Structure of Atoms Pauli Exclusion Principle • No two electrons in the same atom can have exactly the same energy. • Therefore, no two electrons in the same atom can have identical sets of quantum numbers. © 2012 Pearson Education, Inc. Electronic Structure of Atoms Electron Configurations 5 4p • This term shows the distribution of all electrons in an atom. • Each component consists of – A number denoting the energy level, – A letter denoting the type of orbital, – A superscript denoting the number of electrons in those orbitals. Electronic © 2012 Pearson Education, Inc. Structure of Atoms Orbital Diagrams • Each box in the diagram represents one orbital. • Half-arrows represent the electrons. • The direction of the arrow represents the relative spin of the electron. © 2012 Pearson Education, Inc. Electronic Structure of Atoms Hund’s Rule “For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.” © 2012 Pearson Education, Inc. Electronic Structure of Atoms Sample Exercise 6.7 Orbital Diagrams and Electron Configurations Draw the orbital diagram for the electron configuration of oxygen, atomic number 8. How many unpaired electrons does an oxygen atom possess? Solution Practice Exercise (a) Write the electron configuration for phosphorus, element 15. (b) How many unpaired electrons does a phosphorus atom possess? Electronic Structure of Atoms A. 5d B. 6s Electronic Structure of Atoms Periodic Table • We fill orbitals in increasing order of energy. • Different blocks on the periodic table (shaded in different colors in this chart) correspond to different types of orbitals. © 2012 Pearson Education, Inc. Electronic Structure of Atoms Sample Exercise 6.8 Electron Configurations for a Group What is the characteristic valence electron configuration of the group 7A elements, the halogens? Solution Practice Exercise Which family of elements is characterized by an ns2np2 electron configuration in the outermost occupied shell? Electronic Structure of Atoms Sample Exercise 6.9 Electron Configurations from the Periodic Table (a) Based on its position in the periodic table, write the condensed electron configuration for bismuth, element 83. (b) How many unpaired electrons does a bismuth atom have? Solution Practice Exercise Use the periodic table to write the condensed electron configuration for (a) Co (element 27), (b) Te (element 52). Electronic Structure of Atoms Some Anomalies Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row. • • © 2012 Pearson Education, Inc. For instance, the electron configuration for copper is [Ar] 4s1 3d10 rather than the expected [Ar] 4s2 3d9. This occurs because the 4s and 3d orbitals are very close in energy. These anomalies occur in fblock atoms, as well. Electronic Structure of Atoms A. B. C. D All three elements have the same relative energies for their nd and (n + 1)s orbitals. Nothing can be concluded, because the three elements each have different valence electron configurations. The nd orbitals are of lower energy than (n + 1)s orbitals for all three elements. The nd orbitals are of higher energy than Electronic Structure (n + 1)s orbitals for all three elements. of Atoms